3 Answers
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It's not hard to explicitly construct G using the quaternions, assuming P is not \pm Q, and I think this is worth working out in detail because I really like this picture of SU(2). Identify SU(2) with the unit quaternions by the isomorphism

[a b ]
[-\bar{b} \bar{a}] --> a+bj

and since the trace of such a matrix is 2*Re(a), the traceless matrices in SU(2) correspond exactly to the purely imaginary unit quaternions.

Now if we consider R^3 as the space of all imaginary quaternions, we can describe the SU(2) action on it geometrically: Let v=cos(t)+sin(t)w, where w is purely imaginary and |w|=1. Then the conjugation map x -> vxv^{-1} is a rotation of the plane orthogonal to w -- you can easily check that it's in SO(3), and that it fixes w because vw=wv. It's also not hard to check that the angle of this rotation is 2t.

Consider P and Q as imaginary quaternions, hence also as vectors in R^3. Then PQ = -(P.Q) + PxQ, so let G=Im(PQ)/|Im(PQ)|. Now G is a unit vector orthogonal to both P and Q, and conjugation by G is the same as rotating the plane through P and Q by \pi, so GPG^{-1}=-P and GQG^{-1}=-Q as desired.

Finally, we just go back from the unit quaternions to SU(2): up to scale, G was supposed to be the imaginary part of the quaternion PQ, so the matrix G is some constant times the traceless part of the matrix PQ.

Yes. Since det(P)=1 and tr(P)=0, the eigenvalues of P are \pm i. So one of the logarithms of P is p, a matrix in su(2) with eigenvalues i \pi/2 and - i \pi/2. Here su(2) is the Lie algebra of SU(2). (In terms of quaternions, p is of the form bi+cj+dk with b^2+c^2+d^2=(pi/2)^2.) Similarly, Q has a logarithm q in su(2) which also has eigenvalues \pm i pi/2. Let H be the plane in su(2) spanned by p and q.

Now, SU(2) acts on su(2) by the group SO(3) of orthogonal rotations. Let G be the rotation of H by angle pi. So G p G^{-1}=-p and G q G^{-1} = -q. Exponentiating these equations, G has the desired property.